To solve the quadratic equation $2x^2 + 4x + 1 = 0$, we will use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this equation, $a = 2$, $b = 4$, and $c = 1$.

Step 1: Calculate the discriminant

$\Delta = b^2 - 4ac = 4^2 - 4(2)(1) = 16 - 8 = 8$

Step 2: Apply the quadratic formula

$x = \frac{-4 \pm \sqrt{8}}{2(2)} = \frac{-4 \pm 2\sqrt{2}}{4}$

Step 3: Simplify the expression

$x = \frac{-4}{4} \pm \frac{2\sqrt{2}}{4} = -1 \pm \frac{\sqrt{2}}{2}$

Thus, the two solutions are: $x_1 = -1 + \frac{\sqrt{2}}{2}, \quad x_2 = -1 - \frac{\sqrt{2}}{2}$

Let me know if you need further details on any step.

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Here are 5 related questions:

1. How does the discriminant affect the nature of the roots?
2. Can you solve the equation $3x^2 + 5x - 2 = 0$ using the quadratic formula?
3. What happens if the discriminant is negative in a quadratic equation?
4. How can you use completing the square to solve a quadratic equation?
5. Can this equation be factored, or is the quadratic formula necessary?

Tip: The discriminant $\Delta = b^2 - 4ac$ determines whether the roots are real or complex. If $\Delta > 0$, the roots are real and distinct; if $\Delta = 0$, they are real and identical; if $\Delta < 0$, the roots are complex.